Introduction to Optimization

A sheet of canvas $4$ meters wide and $3$ meters long is draped over a horizontal ridge rope along its center line. Each half slopes from the rope to the ground as a rectangle $2$ meters by $3$ meters. The two bottom edges are staked down parallel to the rope, and the ends are left open.

The spacing of the stakes sets the tent's shape. Staked $3.2$ meters apart, the ridge is $1.2$ meters high, and each triangular end has area $\frac{1}{2}(3.2)(1.2) = 1.92$ square meters. Staked $2.4$ meters apart, the ridge is $1.6$ meters high, and the end's area is again $\frac{1}{2}(2.4)(1.6) = 1.92$ square meters. Either way, the tent holds $3(1.92) = 5.76$ cubic meters.

A wide tent is low and a tall tent is narrow, so the volume depends on one choice, the spacing. Finding the spacing with the largest volume is an optimization problem: a quantity is made as large or as small as the situation allows.

The quantity to maximize is the tent's volume $V$, in cubic meters. Let $x$ be half the distance between the staked edges and $h$ the height of the ridge, both in meters. Each end is a triangle with base $2x$ and height $h$, and the tent is $3$ meters long, so

The constraint is the canvas. Each slanted side is $2$ meters long, the hypotenuse of a right triangle with legs $x$ and $h$, so $x^2 + h^2 = 4$. Since the height can't be negative, $h = \sqrt{4 - x^2}$. Substituting it into $V = 3xh$ gives the volume as a function of one variable,

For the domain, $x$ can't be negative, and the half-spacing can't be longer than the $2$-meter side, so $0 \le x \le 2$. Both ends of that interval are flat tents. At $x = 0$ the halves hang straight down against each other, and at $x = 2$ the canvas lies on the ground, so $V = 0$ at both. Keeping them gives the closed interval $[0, 2]$.

The wide tent above has $x = 1.6$, and $V(1.6) = 3(1.6)\sqrt{1.44}$, which is $3(1.6)(1.2) = 5.76$, the volume found there.

The function $V(x) = 3x\sqrt{4 - x^2}$ is continuous on $[0, 2]$ and differentiable on $(0, 2)$. By the product and chain rules,

On $(0, 2)$ the denominator is positive, so $V'(x) = 0$ only where $x^2 = 2$, at $x = \sqrt{2}$. The other root, $x = -\sqrt{2}$, is outside the domain. The derivative is undefined at $x = 2$, which is already a candidate as an endpoint. The candidates are

since $V\left(\sqrt{2}\right) = 3\sqrt{2} \cdot \sqrt{4 - 2} = 3 \cdot 2 = 6$. By the Candidates Test, the absolute maximum of $V$ on $[0, 2]$ is $6$, at $x = \sqrt{2}$. The ridge height is then $h = \sqrt{4 - 2} = \sqrt{2}$ as well.

Stake the bottom edges $2\sqrt{2} \approx 2.83$ meters apart. The ridge is then $\sqrt{2} \approx 1.41$ meters high, and the tent holds $6$ cubic meters, more than the $5.76$ of either tent above. Since $x = h$, each half of the end is an isosceles right triangle, so the two sides of canvas meet at a right angle at the ridge.